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Check if a Binary Tree contains duplicate subtrees of size 2 or more

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Given a Binary Tree, check whether the Binary tree contains a duplicate sub-tree of size 2 or more. 
Note: Two same leaf nodes are not considered as the subtree size of a leaf node is one.

Input :  Binary Tree 

               A

             /    \ 

           B        C

         /   \       \    

        D     E       B     

                     /  \    

                    D    E

Output : Yes


Asked in : Google Interview 


 

Tree


Tree with duplicate Sub-Tree [ highlight by blue color ellipse ]

[ Method 1] 
A simple solution is that, we pick every node of tree and try to find is any sub-tree of given tree is present in tree which is identical with that sub-tree. Here we can use below post to find if a subtree is present anywhere else in tree. 
Check if a binary tree is subtree of another binary tree 

[Method 2 ]( Efficient solution ) 
An Efficient solution based on tree serialization and hashing. The idea is to serialize subtrees as strings and store the strings in hash table. Once we find a serialized tree (which is not a leaf) already existing in hash-table, we return true. 

Below The implementation of above idea. 

C++

// C++ program to find if there is a duplicate
// sub-tree of size 2 or more.
#include<bits/stdc++.h>
using namespace std;
 
// Separator node
const char MARKER = '$';
 
// Structure for a binary tree node
struct Node
{
    char key;
    Node *left, *right;
};
 
// A utility function to create a new node
Node* newNode(char key)
{
    Node* node = new Node;
    node->key = key;
    node->left = node->right = NULL;
    return node;
}
 
unordered_set<string> subtrees;
 
// This function returns empty string if tree
// contains a duplicate subtree of size 2 or more.
string dupSubUtil(Node *root)
{
    string s = "";
 
    // If current node is NULL, return marker
    if (root == NULL)
        return s + MARKER;
 
    // If left subtree has a duplicate subtree.
    string lStr = dupSubUtil(root->left);
    if (lStr.compare(s) == 0)
    return s;
 
    // Do same for right subtree
    string rStr = dupSubUtil(root->right);
    if (rStr.compare(s) == 0)
    return s;
 
    // Serialize current subtree
    s = s + root->key + lStr + rStr;
 
    // If current subtree already exists in hash
    // table. [Note that size of a serialized tree
    // with single node is 3 as it has two marker
    // nodes.
    if (s.length() > 3 &&
        subtrees.find(s) != subtrees.end())
    return "";
 
    subtrees.insert(s);
 
    return s;
}
 
// Driver program to test above functions
int main()
{
    Node *root = newNode('A');
    root->left = newNode('B');
    root->right = newNode('C');
    root->left->left = newNode('D');
    root->left->right = newNode('E');
    root->right->right = newNode('B');
    root->right->right->right = newNode('E');
    root->right->right->left= newNode('D');
 
    string str = dupSubUtil(root);
 
    (str.compare("") == 0) ? cout << " Yes ":
                            cout << " No " ;
    return 0;
}

                    

Java

// Java program to find if there is a duplicate
// sub-tree of size 2 or more.
import java.util.HashSet;
public class Main {
 
    static char MARKER = '$';
 
    // This function returns empty string if tree
    // contains a duplicate subtree of size 2 or more.
    public static String dupSubUtil(Node root, HashSet<String> subtrees)
    {
        String s = "";
     
        // If current node is NULL, return marker
        if (root == null)
            return s + MARKER;
     
        // If left subtree has a duplicate subtree.
        String lStr = dupSubUtil(root.left,subtrees);
        if (lStr.equals(s))
            return s;
     
        // Do same for right subtree
        String rStr = dupSubUtil(root.right,subtrees);
        if (rStr.equals(s))
            return s;
     
        // Serialize current subtree
          // Append random char in between the value to differentiate from 11,1 and 1,11
        s = s + root.data + "%" + lStr+ "%" + rStr;
     
        // If current subtree already exists in hash
        // table. [Note that size of a serialized tree
        // with single node is 7 (3+4 accounting for special chars appended)
        // as it has two marker
        // nodes.
        if (s.length() > 7 && subtrees.contains(s))
            return "";
     
        subtrees.add(s);
        return s;
    }
 
    //Function to find if the Binary Tree contains duplicate
    //subtrees of size 2 or more
    public static String dupSub(Node root)
    {
        HashSet<String> subtrees=new HashSet<>();
        return dupSubUtil(root,subtrees);
    }
 
    public static void main(String args[])
    {
        Node root = new Node('A');
        root.left = new Node('B');
        root.right = new Node('C');
        root.left.left = new Node('D');
        root.left.right = new Node('E');
        root.right.right = new Node('B');
        root.right.right.right = new Node('E');
        root.right.right.left= new Node('D');
        String str = dupSub(root);
        if(str.equals(""))
            System.out.print(" Yes ");
        else   
            System.out.print(" No ");
    }
}
 
// A binary tree Node has data,
// pointer to left child
// and a pointer to right child
class Node {
    int data;
    Node left,right;
    Node(int data)
    {
        this.data=data;
    }
};
//This code is contributed by Gaurav Tiwari

                    

Python3

# Python3 program to find if there is
# a duplicate sub-tree of size 2 or more
 
# Separator node
MARKER = '$'
 
# Structure for a binary tree node
class Node:
     
    def __init__(self, x):
         
        self.key = x
        self.left = None
        self.right = None
 
subtrees = {}
 
# This function returns empty if tree
# contains a duplicate subtree of size
# 2 or more.
def dupSubUtil(root):
     
    global subtrees
 
    s = ""
 
    # If current node is None, return marker
    if (root == None):
        return s + MARKER
 
    # If left subtree has a duplicate subtree.
    lStr = dupSubUtil(root.left)
     
    if (s in lStr):
       return s
 
    # Do same for right subtree
    rStr = dupSubUtil(root.right)
     
    if (s in rStr):
       return s
 
    # Serialize current subtree
    s = s + root.key + lStr + rStr
 
    # If current subtree already exists in hash
    # table. [Note that size of a serialized tree
    # with single node is 3 as it has two marker
    # nodes.
    if (len(s) > 3 and s in subtrees):
       return ""
 
    subtrees[s] = 1
 
    return s
 
# Driver code
if __name__ == '__main__':
     
    root = Node('A')
    root.left = Node('B')
    root.right = Node('C')
    root.left.left = Node('D')
    root.left.right = Node('E')
    root.right.right = Node('B')
    root.right.right.right = Node('E')
    root.right.right.left= Node('D')
 
    str = dupSubUtil(root)
 
    if "" in str:
        print(" Yes ")
    else:
        print(" No ")
 
# This code is contributed by mohit kumar 29

                    

C#

// C# program to find if there is a duplicate
// sub-tree of size 2 or more.
using System;
using System.Collections.Generic;
 
class GFG
{
 
    static char MARKER = '$';
 
    // This function returns empty string if tree
    // contains a duplicate subtree of size 2 or more.
    public static String dupSubUtil(Node root,
                    HashSet<String> subtrees)
    {
        String s = "";
     
        // If current node is NULL, return marker
        if (root == null)
            return s + MARKER;
     
        // If left subtree has a duplicate subtree.
        String lStr = dupSubUtil(root.left,subtrees);
        if (lStr.Equals(s))
            return s;
     
        // Do same for right subtree
        String rStr = dupSubUtil(root.right,subtrees);
        if (rStr.Equals(s))
            return s;
     
        // Serialize current subtree
        s = s + root.data + lStr + rStr;
     
        // If current subtree already exists in hash
        // table. [Note that size of a serialized tree
        // with single node is 3 as it has two marker
        // nodes.
        if (s.Length > 3 && subtrees.Contains(s))
            return "";
     
        subtrees.Add(s);
        return s;
    }
 
    // Function to find if the Binary Tree contains
    // duplicate subtrees of size 2 or more
    public static String dupSub(Node root)
    {
        HashSet<String> subtrees = new HashSet<String>();
        return dupSubUtil(root,subtrees);
    }
 
    // Driver code
    public static void Main(String []args)
    {
        Node root = new Node('A');
        root.left = new Node('B');
        root.right = new Node('C');
        root.left.left = new Node('D');
        root.left.right = new Node('E');
        root.right.right = new Node('B');
        root.right.right.right = new Node('E');
        root.right.right.left= new Node('D');
        String str = dupSub(root);
        if(str.Equals(""))
            Console.Write(" Yes ");
        else
            Console.Write(" No ");
    }
}
 
// A binary tree Node has data,
// pointer to left child
// and a pointer to right child
public class Node
{
    public int data;
    public Node left,right;
    public Node(int data)
    {
        this.data = data;
    }
};
 
// This code is contributed by 29AjayKumar

                    

Javascript

<script>
// Javascript program to find if there is a duplicate
// sub-tree of size 2 or more.
     
    let MARKER = '$';
     
    // A binary tree Node has data,
// pointer to left child
// and a pointer to right child
    class Node {
        constructor(data)
        {
            this.data=data;
        }
    }
     
    // This function returns empty string if tree
    // contains a duplicate subtree of size 2 or more.
    function dupSubUtil(root,subtrees)
    {
        let s = "";
      
        // If current node is NULL, return marker
        if (root == null)
            return s + MARKER;
      
        // If left subtree has a duplicate subtree.
        let lStr = dupSubUtil(root.left,subtrees);
        if (lStr==(s))
            return s;
      
        // Do same for right subtree
        let rStr = dupSubUtil(root.right,subtrees);
        if (rStr==(s))
            return s;
      
        // Serialize current subtree
        s = s + root.data + lStr + rStr;
      
        // If current subtree already exists in hash
        // table. [Note that size of a serialized tree
        // with single node is 3 as it has two marker
        // nodes.
        if (s.length > 3 && subtrees.has(s))
            return "";
      
        subtrees.add(s);
        return s;
    }
     
    //Function to find if the Binary Tree contains duplicate
    //subtrees of size 2 or more
    function dupSub(root)
    {
            let subtrees=new Set();
             
            return dupSubUtil(root,subtrees);
    }
     
    let root = new Node('A');
    root.left = new Node('B');
    root.right = new Node('C');
    root.left.left = new Node('D');
    root.left.right = new Node('E');
    root.right.right = new Node('B');
    root.right.right.right = new Node('E');
    root.right.right.left= new Node('D');
    let str = dupSub(root);
    if(str==(""))
        document.write(" Yes ");
     else  
        document.write(" No ");
     
 
// This code is contributed by unknown2108
</script>

                    

Output
 Yes 

Time Complexity: O(n)
Auxiliary Space: O(n)

The time complexity of the above program is O(n) where n is the number of nodes in the given binary tree. We are using hashing to store the subtrees which take O(n) space for all the subtrees.




 



Last Updated : 27 Jan, 2023
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